How to show the cyclotomic polynomial is irreducible over $\mathbb{R}(T,\sqrt[n]{T})$

Question

I'm trying to solve the following problem.

Let $T$ be a transcendental over $\mathbb{R}$. Put $K=\mathbb{R}(T), n\geq3$. Let $L$ be the least splitting field of $X^n-T$. Then, calculate $[L:K]$.

I think the answer is $n\phi(n)$. Because $L=\mathbb{R}(T,\sqrt[n]{T},\xi)$ where $\xi$ is $n$-th root of unity, and I think $[\mathbb{R}(T,\sqrt[n]{T},\xi):\mathbb{R}(T,\sqrt[n]{T})]=\phi(n), [\mathbb{R}(T,\sqrt[n]{T}):\mathbb{R}(T)]=n$.

The latter is okay because $X^n-T$ is irreducible in $\mathbb{R}(T)[X]$, but I can't show the former.

How can I show the cyclotomic polynomial is irreducible?

Answer 1

I'm afraid you are aiming at the wrong target.

Hint: $\xi$ is a non-real complex number, so $[\Bbb{R}(\xi):\Bbb{R}]=?$